Linear Algebra - Orthogonalization - Building an orthogonal set of generators

Original stated goal: Find the projection of b orthogonal to the space V spanned by arbitrary vectors <math>{v_1} , \dots , {v_n}</math>

• Input: A list <math>{v_1} , \dots , {v_n}</math> of vectors over the reals
• output: A list of mutually orthogonal vectors <math>v_1^* , \dots , v_n^*</math> such that

<MATH>Span \{ v_1^* , \dots , v_n^* \} = Span \{v_1, \dots , v_n\}</MATH>

# output: a list of mutually orthogonal vectors.
def orthogonalize(vlist):
    vstarlist = []
    for v in vlist:
        # project orthogonal find the next orthogonal vector
        # and make iteratively a longer and longer list of mutually orthogonal vectors.
        vstarlist.append(project_orthogonal(v, vstarlist))
    return vstarlist

where:

• the project orthogonal function is the high dimensional projection computation of mutually orthogonal vectors.

Lemma: Throughout the execution of orthogonalize, the vectors in vstarlist are mutually orthogonal.

Proof: by induction, using the fact that each vector added to vstarlist is orthogonal to all the vectors already in the list.

If we run the procedure orthogonalize twice, once with a list of vectors and once with the reverse of that list, the output lists will not be the reverses of each other.

• First iteration, since the set <math>\{v_0\}</math> is trivially a set of mutually orthogonal vectors:
  • <math>v^*_0 = v_0</math>
  • or as matrix equation: <math>[v_1] = [v^*_1] [1]</math>
• Second Iteration, <math>v^*_1</math> is the the projection orthogonal of <math>v_1</math> to <math>v^*_0</math> then:
  • <math>v_1 = v^*_1 + \sigma_{01} v^*_0</math>
  • or as matrix equation: <math>[v_1] = \begin{bmatrix}v^*_0 & v^*_1 \end{bmatrix} \begin{bmatrix}\sigma_{01} \\ 1 \end{bmatrix}</math>
• Third iteration: <math>v^*_2</math> is the projection of <math>v_2</math> orthogonal to <math>v^*_0</math> and <math>v^*_1</math> :
  • <math>v_2 = v^*_2 + (\sigma_{02} v^*_0 + \sigma_{12} v^*_1)</math>
  • or as matrix equation: <math>[v_2] = \begin{bmatrix}v^*_0 & v^*_1 & v^*_2 \end{bmatrix} \begin{bmatrix}\sigma_{02} \\ \sigma_{12} \\ 1 \end{bmatrix}</math>
• After N Iterations, we will get this matrix equation (formed of two matrix) expressing original vectors in terms of starred vectors. <note warning>It's not a matrix multiplication</note>

<MATH> \begin{bmatrix} \begin{array}{r|r|r|r} \, \: \; \> & \\ \, \: \; \> & \\ \\ v_0 & v_1 & v_2 & \dots & v_n \ \\ \ \\ \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{r|r|r|r} v^*_0 & v^*_1 & v^*_2 & \dots & v^*_n \end{array} \end{bmatrix} \begin{bmatrix} 1 & \sigma_{01} & \sigma_{02} & \dots & \sigma_{0n} \\ & 1 & \sigma_{12} & \dots & \sigma_{1n} \\ & & 1 & \dots & \sigma_{2n} \\ & & & \ddots & \vdots \\ & & & & 1 \\ \end{bmatrix} </MATH>

The two matrices on the right are special:

• Columns of first one are mutually orthogonal.
• Second is upper triangular.